Kinematics-03-SUBJECTIVE SOLVED

SOLVED PROBLEMS 1. Position of a particle moving along x-axis is given by x  3t  4t 2  t 3 , where x is in meters and t in seconds. (a) Find the position of the particle at t  2s . (b) Find the displacement of the particle in the time interval from t  0 to t  4s . (c) Find the average velocity of the particle in the time interval from t  2s to t  4s . (d) Find the velocity of the particle at t  2s . Solution: (a)  (b) x(t)  3t  4t 2  t3 x  3  2  4  (2)2  (2)3  6  4  4  8  2m . x (0)  0 x (4)  3  4  4  (4) 2  (4)3  12m . Displacement  x (4)  x (0)  12m (c) (d)  v  x (4)  x (2) (4  2) dx  3  8t  3t 2 dt  12  (2) m / s  7 m / s 2  v(2)  dx   dt   3  8  2  3  (2)2  1 m / s  (2) 2. A body moving in a curved path possesses a velocity 3 m/s towards north at any instant of its motion. After 10s, the velocity of the body was found to be 4 m/s towards west. Calculate the average accel- eration during this interval. Solution: To solve this problem the vector nature of velocity must be taken into account. In the figure, the initial velocity v0 and the final velocity v are drawn from a common origin. The vector difference of them is found by the parallelogram method. The magnitude of difference is N v  v0  OC    5 m / s The direction is given by W E tan   3  0.75 ,   37º  Average acceleration → →  | v  v0 |  5 S t 10 = 0.5 m/s2 at 37º South of West. 3. The velocity time graph of a moving object is given in the figure. Find the maximum acceleration of the body and distance travelled by the body in the interval of time in which this acceleration exists. Solution: 80 60 40 20 10 20 30 40 50 60 70 80 time(in sec.) Acceleration is maximum when slope is maximum. amax  80  20  6 m / s 2 40  30 Displacement is the area under the curve.  S  1 (20  80)(10)  500m 2 4. A monkey is climbing a vertical tree with a velocity of 10 m/s while a dog runs towards the tree chasing the monkey with a velocity of 15 m/s. Find the velocity of the dog relative to the monkey. Solution: Velocities are shown in the figure Velocity of the dog relative to the monkey = velocity of Dog – velocity of monkey → →  VD  VM → →  O VD  (VM ) This velocity is directed along OM and its magnitude is D  VM   18 m/s M This velocity makes an angle  with the horizontal, where tan   10  2 or   tan1 2   33.7º 15 3    3  5. A steel ball is dropped from the roof of a building. An observer standing in front of a window 1.5m high 1 notes that the ball takes 10 s to fall from the top to the bottom of the window. The ball reappears at the bottom of the window 2s after passing it on the way down. If the collision between the ball and the ground is perfectly elastic, then find the height of the building. Take g = 10 m/s2. Solution: Since collision is perfectly elastic, the speed of the ball just before collision is equal to the speed of the ball just after collision. Hence time of descent is equal to the time of ascent. Therefore time taken by the ball to reach the ground from the bottom of the window is 1 sec. Let u be the speed of the ball when it is at the top of the window 1 1 1  → → 1 → 2   1.5  u 10  2 10  100 .  s  ut  at   2   u = 14.5 m/s  ball is dropped hence its initial speed is 0 It t be the time taken by the ball to acquire the speed of 14.5 m/s, then 14.5  0 10  t, (v  u  at)  t = 1.45 sec Hence total time of descent is given by T  1.45  1 10  1  2.55s If H be the height of the building, then H  0  1 gt 2 2  H  1 10  (2.55)2 2  H  32.5 m 6. A block of ice starts sliding down from the top of an inclined roof of a house along a line of the greatest slope. The inclination of the roof with the horizontal is 30º. The heights of the highest and lowest points of the roof are 8.1 m and 5.6 m respectively. At what horizontal distance from the lowest point will the block hit the ground? Neglect any friction. [g = 9.8 m/s2] Solution: Acceleration of the block along the greatest slope is equal to a = g sin 30º Distance travelled by the block along the greatest slope is equal to S  (8.1  5.6)  5m . sin 30º If u be the speed of the block when it is just about to leave the roof then u 2  0  2g sin 30º5 u = 7 m/s If t be the time taken to hit the ground then 5.6  u sin 30º t   7t 2  5t  8  0 1 gt 2  2 7 t  2 1  9.8t 2 2 8.1 t  2  7  t   5  15.78 s , -ve value is to be rejected. 14 i.e., t   5  15.78  0.77 14 sec. Horizontal distance travelled is equalto x  u cos 30º t  7  x  4.67 m 3  10.78 m 2 14 7. A man wants to cross a river 500 m wide. The rowing speed of the man relative to water is 3 km/hr and the river flows at the speed of 2 km/hr. If the man’s walking speed on the shore is 5 km/hr, then in which direction should he start rowing in order to reach the directly opposite point on the other bank in the shortest time. Solution: Lets → velocity of the man relative to ground. v = velocity of the man relative to water B u = velocity of water, y given | u | 2kmh → | v |  3kmh1 x Let him start at an angle  with the normal u Net velcoity of man → → u  v A ˆ ˆ vm  (u  vsin )i  v cos j Hence time taken by the man to cross the river is  Drift of the man along the river is t1  0.5 v cos  x  (u  v sin )t1 0.5 x  (u  v sin ) v cos  Time taken by the man to cover this distance is  u sec  0.5  tan  t   v    u sec   tan  2 0.1  5  v  Therefore, total time T  t1  t 2  T  0.5 sec   0.1u sec   0.1 tan  v v Putting the value of u and v, we get T  0.5 sec   0.1  2 sec   0.1 tan  3 3  0.7 sec   0.1tan  3  dT  0.7 sec  tan   – 0.1 sec 2  d 3 for T to be minimum , dT  0 d  sin   (3 / 7)    sin1(3 / 7) 8. Two particles A and B move with constant velocities v1 and v 2 along two mutually perpendicular straight lines towards the intersection point O. At moment t = 0, the particle were located at distance l1 and l 2 from O, respectively. Find the time, when they are nearest and also the shortest distance between them. Solution: Method I: → → → ˆ ˆ  v AB  v A  v B  v1i  v 2 j Minimum distance is the length of the perpendicular to → from B. If  is the angle between the x – axis and → , then tan     v 2 v1 v 2 In AOD, OD  OA tan   l1 x 1 Therefore BD  l  OD  v1l 2  v 2 l1 l2 2 v In BCD, 1 cos   BC BD BC  BD cos   v1l 2  v 2 l1   v1  BC  | v1l 2  v 2 l1 | AC The required time → AD  DC → v AB v AB l1 v2  v2  v1l2  v2l1  v2 l se an  v 1 2 v2  v2 v l v  l v  1  1 1 2 1  1 1 2 2 . Method II: v2  v2 v2  v2 After time ‘t’, the position of the point A and B are (l1  v1t) The distance L between the points A’ and B’ are and (l2  v2 t) , respectively. L2  (l1  v1t)2  (l2  v 2 t)2 Differentiating with respect to time, ...(i) 2L dL  2(l  v t)(v )  2(l  v t)(v ) From minimum value of L, dL  0 dt 1 1 1 2 2 2 dt (v 2  v 2 )t  l v  l v . 1 2 1 1 2 2 l1 l v  l v A O 1 1 2 2 Or v2  v 2 1 2 Putting the value of t in equation (1) l2 B' v2 Lmin  . 9. A particle moving with uniform acceleration in a straight line covers a distance of 3 m in the 8th second and 5 m in the 16th second of its motion. What is the displacement of the particle from the beginning of the 6th second to the end of 15th second? Solution: The distance traveled during the nth second of motion of a body is given by S  u  1 a(2n  1) n 2 S  un  1 an2 2 For the motion during the 8th second, 3  u  1 a(16  1)  u  15a ...(i) u(n 1)  1 a(n 1)2 2 2 2 For the motion during the 16th second, 5  u  1 a(32  1)  u  31a 2 2 Subtracting equations (i) from (ii) 8a  2 or acceleration a  1 ms2 4 From equation (i), u  3   15  1   9 ms1     Now, the velocity at the end of 5 s (velocity at the beginning of 6th second) v1  u  5a The velocity at the end of 15th s, v 2  u  15a Average velocity during this interval of 10 seconds  (u  5a)  (u  15a)  u  10a 2 Distance travelled during this interval S = average velocity x time  (u  10a)  t   9  10 10  290  36.25m .  v1  v 2 2     10. An automobile can accelerate or decelerate at a maximum value of 5 m / s 2 3 and can attain a maximum speed of 90 km/hr. If it starts from rest, what is the shortest time in which can travel one kilometre, if it is to come to rest at the end of kilometre run? Solution: In order that the time of motion be shortest, the car should attain the maximum velocity with the maximum acceleration after the start, maintain the maximum velocity for as along as possible and then decelerate with the maximum retardation possible, consistent with the condition that, the automobile should come to rest immediately after covering a distance of 1 km. Let t1 be the time of acceleration, t 2 be the time of uniform velocity and t 3 be the time of retardation. Now, maximum velocity possible = 90 km/hr  90  5 18 m/s = 25 m/s t  v  u  25  0  15s 1 a 5 3 Similarly, the me of retardation is also given by t  0  25  15s 3  5 3 During the period of acceleration, the distance covered = average velocity  time  25  0 15  187.5m 2 During the period of deacceleration, the distance covered is the same the hence = 187.5 m the total distance covered under constant velocity = 1000 – 375 = 625 m Time of motion under constant velocity. t 2  625  25s 25 the shortest time of motion  t1  t 2  t 3  15  25  15  55 seconds. 11. Two particles are projected at the same instant from two points A and B on the same horizontal level where AB = 28 m, the motion taking place in a vertical plane through AB. The particle from A has an initial velocity of 39 m/s at an angle sin 1  5  13 with AB and the particle from B has an initial velocity   sin 1  3  of 25 m/s at an angle  5  with BA. Show that the particle would collide in mid-air find when and   where the impact occurs. Solution: AB = 28m. At A, a particle is projected with velocity u = 39 m/s. u1 and u 2 are its horizontal and vertical compo- nents respectively. The angle u makes with AB cos 1 . Given that sin 1  13  cos 1  12 . 13 Similarly for the particle projected from B, with velocity v =25 m/s, v1 and v 2 are the horizontal and vertical components respectively. sin  2  5  cos  2 5  4 . 5 u2 v2 Now u 2  u sin 1  39  13  15 m/s A u1 28m v1 B v 2  v sin  2  25  3  15 5 m/s The vertical components of the velocities are the same at the start. Subsequently at the other instant t their vertical displacement are equal and have a value h  15t  4.9t 2 which means that the line joining their positions at the instant t continues to be horizontal and the particles come closer to each other. Their relative velocity in the horizontal direction = 39 cos 1  25 cos  2  39  12  25  4  36  20  56 m / s 13 5 Time of collision  AB  28  0.5s , after they were projected. 56 56 Height at which the collision occurs  ut  1 at 2  15(0.5)  1 (9.8)(0.5)2 2 2 = 6.275 m The horizontal distance of the position of collision from A  (u1 cos1) (time of collision)  39  12  0.5s 13  18 m 12. A man can swim with a velocity V1 relative to water in a river flowing with speed V2 .Show that it will take him times as long to swim a certain distance upstream and back as to swim the same distance and back perpendicular to the direction of the stream (V1  V2 ). Solution: Suppose the man swims a distance x up and the same distance down the stream. Velocity of man upstream relative to the ground  V1  V2 . Time taken for this, t1  1 x  V2 Velocity of man downstream relative to the ground  V1  V2 Time taken for this, t 2  x V1  V2 Total time taken t1  t 2  V  V  x V  V  2V1x V 2  V 2 1 2 1 2 1 2 Next the man intends crossing the river perpendicular to the direction of the stream. If he wants to cross the river straight across he must swim in a direction OM such that the vector sum of velocity of man + velocity of river will give him a velocity relative to the ground in a direction perpendicular to the direction of the stream. In the Figure the velocity relative to the ground is OR and the magnitude of OR  Now the man swims a distance x up and x down perpendicular to the river flow. Time taken for this, t  2x t 2V1x  t (V2  V2 ) M V2 V R V  V Then the ratio, 1 2  1 2 1  1 2 t 2x O 2V x V 2  V 2 V  1  1 2  1 . V 2  V 2 2x 1 2 13. A particle thrown over a triangle from one end of a horizontal base of a vertical triangle gazes the vertex and falls on the other end of the base. If  and  be the base angle and  the angle of projection, prove that tan  = tan  tan . Solution: The situation is shown in the figure. From figure, Y P(x, y) tan   tan  y x  y (R  x) u    where R is the range. O x B (R-x) A X tan   tan  y(R  x)  xy x(R  x) or tan   tan  y  x R (R  x) ...(i) we known y  x tan 1  x     R  or tan   y  R ...(ii) x (R  x) from equation (i) and (ii), we have tan   tan   tan . 14. A batsman hits a ball at a height of 1.22 m above the ground so that ball leaves the bat an angle 45º with the horizontal. A 7.31 m high wall is situated at a distance of 97.53 m from the position of the batsman. Will the ball clear the wall if its maximum horizontal distance from point of projection is 106.68 m. Take g = 10 m/s2. Solution: v 2 sin 2 v0 R(range)  0 g 2 Rg  v 0  sin 2  Rg as   45º 106.68m  v 0  Rg Equation of trajectory Y  x tan 45º gx 2 2v 2 cos 2 45º  x  gx 2 2Rg 1 2 using (1) Putting x = 97.53, we get 10  (97.53)2 y  97.53  106.68 10  8.35 Hence height of the ball from the ground level is h = 8.35 + 1.22 = 9.577 m. as height of the wall is 7.31 m so the ball will clear the wall. 15. A particle projected with velocity v 0 , strikes at right angles a plane through the point of projection and of inclination  with the horizontal. Find the height of the point struck, from horizontal plane through the point of projection. Solution: Let  be the angle between the velocity of projection and the inclined plane. v 0x  v 0 cos , v 0y  v 0 sin  a x  g sin  a y  g cos   vx (t)  v 0 cos   g sin t At the point of impact v x  0 t  v 0 cos  g sin  ...(i) Also y at the point is zero.  v 0 sin t  1 g cos t 2  0 2 t  2v 0 sin  g cos ...(ii) From (i) and (ii) v0 cos  2v0 sin  gsin g cos tan   1 cot ...(iii) 2 x  v0 cos(  )t  v 0 cos  cos   sin  sin  v 0 cos  g sin  v2 0 cos 2 g  cot   sin  cos  v2  2 2 cot 2   0   cot   (using tan   1 cot ) g  4  cot2   4  cot2  4  cot 2    v2    2 2 cot  g 4  cot 2  From figure v2 2 cot  2v2 y  x tan  0 . g 4  cot 2  tan y  0 g(4  cot 2 ) 16. A ball is thrown from the origin in the x – y plane with velocity 28.28 m/s at an angle 45º to the x-axis. At the same instant a trolley also starts moving with uniform velocity of 10 m/s along the positive x-axis. Initially rear end of the trolley is located at 38m from the origin. Determine the time and position at which the ball hits the trolley. Solution: Let t be the instant at which the ball hits rear face AB of the trolley. Then Or (v0 cos 45  u 0 )t  38 t  38  38  3.8s v 0 cos 45ºu 0 28.28 cos 45º10 At t  3.8s , the y – coordinate of the ball is y  (v 0 sin 45º )t  1 gt 2  20t  5t 2 2 or y  20(3.8)  5(3.8)2  3.8 m Since 3.8 m > 2m, therefore, the ball cannot hit the rear face of the trolley. Now, we assume that the ball hits the top face BC of the trolley, and let t’ be that instant. Then, y  2  20t  5t2 or t 2  4t  0.4  0 t   3.9s Let d be the distance from the point B at which the ball hits the trolley. Then, d  (v 0 cos 45ºu 0 )(t  t)  (20  10)(3.9  3.8)  1m 17. Find the radius of a rotating wheel if the linear velocity v1 of a point on the rim is 2.5 times greater than the linear velocity v 2 of a point 5 cm closer to wheel axle. Solution: Let the radius of the disc = r (in cm.)  v 2  (r  5) v1 r v1 is 2.5 times greater than v 2 v 2  2.5v 2  r  5 r  r  2.5 r  12.5  1.5r  12.5  r  12.5  8.33 1.5 cm. 18. A particle is revolving with a constant angular acceleration  in a circular path of radius r. Find the time when the centripetal acceleration will be numerically equal to the tangential acceleration. Solution: Le the speed of the particle after time t from starting be v  The centripetal acceleration v 2 a r  r  r2 & the corresponding angular speed   t .  a r  r(t)2  r 2 t 2 ...(i) We known that the tangential acceleration a t  r ...(ii) Since, a r  a t (given)  r 2 t 2  r  t  1 .